# Differential as bundle map and pull back bundle

In the wikipedia article about the pushforward, it is stated that if $$f: M\to N$$ is smooth, then it induces a bundle map $$df: TM \to TN$$. It is then claimed that equivalently $$f_*=df$$ is a bundle map from $$TM$$ to the pullback bundle $$f^* TN$$.

Why is this equivalent? The bundles $$TN$$ and $$f^* TN$$ are clearly not the same as they are bundles over different spaces. They could be isomorphic, but is still seems strange that this would hold independently of $$f$$.

Edit: The full quote is

Equivalently (see bundle map), φ∗ = dφ is a bundle map from TM to the pullback bundle φ∗TN over M, which may in turn be viewed as a section of the vector bundle Hom(TM, φ∗TN) over M. The bundle map dφ is also denoted by Tφ and called the tangent map. In this way, T is a functor.

• $TM$ and $f^*TN$ are both bundles over $M$, not bundles over different spaces. – user10354138 Jun 11 at 15:07
• @user10354138 You are right. But whis was only a typo. As can be seen from the context of the question, I am interested in $TN$ and $f^*TN$. I corrected the question accordingly – Michael Jun 11 at 15:41
• Please, provide a complete quote, afik, the wikipedia article contains no such claim. (It says, however, correctly, that "Then, applying the differential pointwise to $X$ yields the pushforward $\phi_*X$, which is a vector field along $\phi$, i.e., a section of $\phi^*TN$ over $M$." The word "yields" is not the same as "is". – Moishe Kohan Jun 11 at 15:55
• @MoisheKohan: I added the quote – Michael Jun 12 at 11:52
• OK, then it is just sloppyness in the article; keep in mind that different parts could have been written by different people. The correct statement is "Then, applying the differential pointwise ..." – Moishe Kohan Jun 12 at 13:52

Recall the definition of the pull-back bundle (in the context you are interested in): If $$f: M\to N$$ is a smooth map, $$df: TM\to TN$$ is the differential of $$f$$, then (as a topological space) $$f^*(TN)=\{(x,\xi): x\in M, \xi\in TN, \xi\in T_{f(x)}N\}\subset M\times TN.$$ The bundle structure on $$f^*(TN)$$ is given by the projection $$\pi: (x,\xi)\mapsto x.$$ Now, $$df$$ defines a bundle map $$TM\to f^*(TN)$$ which I will denote $$Df$$: $$Df: (x,\eta)\mapsto (x, df_x(\eta)), x\in M, \eta\in T_xM.$$ I will leave it to you to check that $$Df$$ is indeed a morphism of vector bundles $$Df: TM\to f^*(TN).$$ By abuse of notation, one frequently denotes the morphism $$Df$$ simply $$df$$ since the"interesting" parts of these morphisms are the same.
Given a vector field $$X\in {\mathfrak X}(M)$$, the push-forward $$f_*(X)$$ is the section $$Y$$ of $$f^*(TN)$$ which is given by the formula $$Y_x= Df(X_x).$$ That's all what there is to it. The wikipedia article commits a minor and common abuse of notation by using the notation $$df$$ for $$Df$$.
• Thank you, this written very clearly. Just one last thing: If $f$ is a submersion, does this mean $TN \simeq f^* TM$? – Michael Jun 14 at 9:33
• Really? What about if $f: X\times Y\to X$ is a projection? – Michael Jun 15 at 10:40
• Yes, really. As I said, you need the equality of ranks of $TN$ and $f^*TM$: It is clearly a necessary condition for isomorphism of these bundles. Note that the bundles $f^*TM$ and $TM$ have equal rank. But rank of the tangent bundle equals the dimension of the manifold. Hence, you are asking for $dim(M)=dim(N)$. But if you have a submersion between manifolds of equal dimension, it has to be a local diffeomorphism. – Moishe Kohan Jun 15 at 13:29